You have found the following ages (in years) of all 6 gorillas at your local zoo: $ 22,\enspace 27,\enspace 13,\enspace 1,\enspace 19,\enspace 22$ What is the average age of the gorillas at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Explanation: Because we have data for all 6 gorillas at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $6$ ages and divide by $6$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\mu} = \dfrac{22 + 27 + 13 + 1 + 19 + 22}{{6}} = {17.3\text{ years old}} $ Find the squared deviations from the mean for each gorilla. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $22$ years $4.7$ years $22.09$ years $^2$ $27$ years $9.7$ years $94.09$ years $^2$ $13$ years $-4.3$ years $18.49$ years $^2$ $1$ year $-16.3$ years $265.69$ years $^2$ $19$ years $1.7$ years $2.89$ years $^2$ $22$ years $4.7$ years $22.09$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{22.09} + {94.09} + {18.49} + {265.69} + {2.89} + {22.09}} {{6}} $ $ {\sigma^2} = \dfrac{{425.34}}{{6}} = {70.89\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{70.89\text{ years}^2}} = {8.4\text{ years}} $ The average gorilla at the zoo is 17.3 years old. There is a standard deviation of 8.4 years.